X+ 3y = 13
(y - 3) + 3y =13
4y - 3 = 13
4y -3 + 3 = 13+3
4y = 16
4y/4 = 16/4
y=4
x = 4 - 3
x = 1
A) (1, 4)
Check x+3y =13
1+ 3(4) =13
1 + 12=13
13=13
PhotosRemaining(x) = 745 pictures - (20 pictures/day)*(x), for x≥0
Answer:
the fraction of players are not stretching regularly is 1 ÷ 5
Step-by-step explanation:
The computation of the fraction of the players not stretching regularly is as follows;
Given that
There are total number of 60 footballers
Out of which 48 strech regularly
So, the players who are not streching properly is
= 60 - 48
= 12
Now the fraction is
= 12 ÷ 60
= 1 ÷ 5
hence, the fraction of players are not stretching regularly is 1 ÷ 5
The probability that all three people surveyed is an independent probability and the calculated probability is 0.1225
The probability a surveyed driver is worried = 35%
- Number of people surveyed = 3
Since the probability of being worried is independent ;
- Probability = P(A) × P(B) × P(C)
The probability that all 3 surveyed are worried :
- P(worried) × P(worried) × P(worried)
- (0.35 × 0.35 × 0.35) = 0.1225
Therefore, the probability that all three people surveyed are worried is : 0.1225
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